Question

Two identical conducting spheres, fixed in place, attract each other with an electrostatic force of 0.0900 N when their center-to-center separation is 48.5 cm. The spheres are then connected by a thin conducting wire. When the wire is removed, the spheres repel each other with an electrostatic force of 0.0528 N. Of the initial charges on the spheres, with a positive net charge, what was (a) the negative charge on one of them and (b) the positive charge on the other? (Assume the negative charge has smaller magnitude.)

Answer 1

Answer:

(a)$-0.75\mu$ Coulomb

(b) $3.1 \mu$ Coulomb

Explanation:

Let $q_1$ be the positive charge, in Coulumb, on the one sphere and $q_2$ be the negative charge, in Coulumb, on the other sphere, where $q_{1} > q_2$.

The center-to-center distance between the spheres is,

$d=48.5 cm=0.485 m$.

From Coulomb's law, the magnitude of the force, $F$, between two point charges having magnitudes $q_1 \& q_2$, separated by distance,$d$, is

$F=\frac {1}{4\pi\epsilon_0}\frac{q_1q_2}{d^2}\;\cdots(i)$

where, $\epsilon_0$ is the permittivity of free space and

$\frac {1}{4\pi\epsilon_0}=9\times 10^9$ in SI units.

So, using equation (i), the attractive force between both the spheres is

$9\times 10^9\frac{q_1q_2}{(0.485)^2}=0.09$

$\Rightarrow q_1q_2=2.35225\times 10^{-12}$

$\Rightarrow q_1q_2= 2.35225 \mu^2$ [as $\mu=10^{-6$]

$\Rightarrow q_2=\frac{\mu^2 2.35225}{q_1}\;\cdots (ii)$

When both the sphere is connected by a thin conducting wire, then redistribution of charge take place and net charge , on combining both the spheres, is $q_1-q_2$ ( as $q_1>q_2$).

Finally, the charge density on both the sphere will be the same.

Given that, both the sphere are identical, so, the quantity of charges on of charged on both the conducting spheres, after removing the conduction wire, will be the same.

So, the net charge on the individual sphere is

$\frac {q_1-q_2}{2}$ which is positive in nature.

From equation (i), the repulsive force between the spheres is

$9\times 10^9\times \frac{(q_1-q_2)(q_1-q_2)}{4\times 0.485^2}=0.0528$

$\Rightarrow (q_1-q_2)^2=5.52\mu^2$ [ as $\mu=10^{-6}$]

$\Rightarrow q_1-q_2=2.35\mu\;\cdots(iii)$

$\Rightarrow q_1 - \frac{\mu^2 2.35225}{q_1}= 2.35\mu$ [using equation (ii)]

$\Rightarrow q_1^2- 2.35\mu q_1- 2.35225\mu^2=0$

$\Rightarrow q_1=3.1 \mu$ or $-0.76 \mu$

Taking positive sign as $q_1$ is the magnitude of charge.

So, for $q1=3.1 \mu$

$q_2=3.1- \mu- 2.35\mu=0.75 \mu$ [from equation (iii)]

Hence,

(a) Negative charge on one of the spheres is,

$q_2=-0.75 \mu$ Coulomb.

(b) Positive charge on the other sphere is,

$q_1=3.1 \mu$ Coulomb.